Contingency Tables and Tests of Association

A contingency table (also called a two-way table) displays the frequency distribution of two categorical variables. The chi-squared test of association determines whether the two variables are independent.

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Structure of a Contingency Table

An $r \times c$r×c contingency table has $r$r rows and $c$c columns:

	Column 1	Column 2	$\cdots$⋯	Column $c$c	Row Total
Row 1	$O_{11}$O11​	$O_{12}$O12​	$\cdots$⋯	$O_{1c}$O1c​	$R_1$R1​
Row 2	$O_{21}$O21​	$O_{22}$O22​	$\cdots$⋯	$O_{2c}$O2c​	$R_2$R2​
$\vdots$⋮	$\vdots$⋮	$\vdots$⋮		$\vdots$⋮	$\vdots$⋮
Row $r$r	$O_{r1}$Or1​	$O_{r2}$Or2​	$\cdots$⋯	$O_{rc}$Orc​	$R_r$Rr​
Col Total	$C_1$C1​	$C_2$C2​	$\cdots$⋯	$C_c$Cc​	$N$N

where $N$N is the grand total.

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Hypotheses

$H_0$H0​: There is no association between the two variables (they are independent).

$H_1$H1​: There is an association between the two variables (they are not independent).

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Expected Frequencies

Under $H_0$H0​ (independence), the expected frequency for cell $(i, j)$(i,j) is:

$E_{ij} = \frac{R_i \times C_j}{N}$Eij​=NRi​×Cj​​

This follows from the multiplication rule for independent events: $P(\text{row } i \cap \text{col } j) = P(\text{row } i) \times P(\text{col } j)$P(row i∩col j)=P(row i)×P(col j).

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The Test Statistic

$X^2 = \sum_{\text{all cells}} \frac{(O_{ij} - E_{ij})^2}{E_{ij}}$X2=∑all cells​Eij​(Oij​−Eij​)2​

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Degrees of Freedom

$\nu = (r - 1)(c - 1)$ν=(r−1)(c−1)

Table size	$\nu$ν
$2 \times 2$2×2	1
$2 \times 3$2×3	2
$3 \times 3$3×3	4
$3 \times 4$3×4	6

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Worked Example: 2 x 3 Table

A survey asks 300 students whether they prefer subject A, B, or C, categorised by gender:

	Subject A	Subject B	Subject C	Row Total
Male	60	40	50	150
Female	30	50	70	150
Col Total	90	90	120	300

$H_0$H0​: Gender and subject preference are independent.

Expected frequencies: $E_{ij} = \frac{R_i \times C_j}{300}$Eij​=300Ri​×Cj​​

	Subject A	Subject B	Subject C
Male	$\frac{150 \times 90}{300} = 45$300150×90​=45	$\frac{150 \times 90}{300} = 45$300150×90​=45	$\frac{150 \times 120}{300} = 60$300150×120​=60
Female	45	45	60

Test statistic:

Cell	$O$O	$E$E	$(O-E)^2/E$(O−E)2/E
M, A	60	45	5.000
M, B	40	45	0.556
M, C	50	60	1.667
F, A	30	45	5.000
F, B	50	45	0.556
F, C	70	60	1.667

$X^2 = 5.000 + 0.556 + 1.667 + 5.000 + 0.556 + 1.667 = 14.446$X2=5.000+0.556+1.667+5.000+0.556+1.667=14.446

Degrees of freedom: $\nu = (2-1)(3-1) = 2$ν=(2−1)(3−1)=2.

At 5%, $\chi^2_2 = 5.991$χ22​=5.991.